Answer
The correct answer is (c).
$2A + B_2 --\gt A_2B_2$
Work Step by Step
** Since there are 2 blue spheres together, we can call them $B_2$.
If we count the spheres, we can see that, in the left box, there are 10 "A"s, and 5 "$B_2$"s. Doing the same thing for the right box, we can count 5 "$A_2B_2$".
Writing this:
$10A + 5B_2 --\gt 5A_2B_2$
Simplifying gives:
$\frac{10}{5}A + \frac{5}{5}B_2 --\gt \frac{5}{5}A_2B_2$
$2A + B_2 -- \gt A_2B_2$
